2-Find-the-term-indepen-dent-of-x-in-the-expansion-of-2x-2-1-x-6-

Question Number 27882 by das47955@mail.com last updated on 16/Jan/18

(2) Find the term indepen -

dent of x in the expansion of

{(2 x^{2} + \frac{1}{x})}^{6}

Answered by Rasheed.Sindhi last updated on 16/Jan/18

T_{r + 1} = (\begin{matrix} n \\ r \end{matrix}) a^{n - r} b^{r}

T_{r + 1} = (\begin{matrix} 6 \\ r \end{matrix}) {({2 x}^{2})}^{6 - r} {(\frac{1}{x})}^{r}

= (\begin{matrix} 6 \\ r \end{matrix}) {(2)}^{6 - r} (x^{12 - 2 r}) (\frac{1}{x^{r}})

= (\begin{matrix} 6 \\ r \end{matrix}) {(2)}^{6 - r} (x^{12 - 3 r})

If T_{r + 1} is free of x

12 - 3 r = 0 \Rightarrow r = 4

T_{r + 1} = T_{4 + 1} = T_{5}

T_{5} is free of x